算法学习(5):ST表
2021/4/30 20:28:19
本文主要是介绍算法学习(5):ST表,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
ST表(Sparse Table,稀疏表)是一种简单的数据结构,主要用来解决RMQ(Range Maximum/Minimum Query,区间最大/最小值查询)问题。
//进行预处理,计算区间最大值 int f[MAXN][21]; // 第二维的大小根据数据范围决定,不小于log(MAXN) for (int i = 1; i <= n; ++i) f[i][0] = read(); // 读入数据 for (int i = 1; i <= 20; ++i) for (int j = 1; j + (1 << i) - 1 <= n; ++j) f[j][i] = max(f[j][i - 1], f[j + (1 << (i - 1))][i - 1]); //预处理log计算 for (int i = 2; i <= n; ++i) Log2[i] = Log2[i / 2] + 1; //查询 for (int i = 0; i < m; ++i) { int l = read(), r = read(); int s = Log2[r - l + 1]; printf("%d\n", max(f[l][s], f[r - (1 << s) + 1][s])); }
例题:
(洛谷P2880 [USACO07JAN]平衡的阵容Balanced Lineup)
For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.
Farmer John has made a list of Q (1 ≤ Q ≤ 180,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.
input:
Line 1: Two space-separated integers, N and Q.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2..N+Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.
output:
Lines 1..Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.
#include<bits/stdc++.h> using namespace std; int Log[50000+2],Max[50000+2][20],Min[50000+2][20]; int main() { int n,m; cin >> n >> m; for(int i = 1;i <= n; ++ i) { cin >> Max[i][0]; Min[i][0] = Max[i][0]; } for(int i = 2;i <= n; ++ i) { Log[i] = Log[i / 2] + 1; } for(int i = 1;i < 20; ++ i) { for(int j = 1;j + (1 << i) - 1 <= n; ++ j) { Min[j][i] = min(Min[j][i - 1], Min[j + (1 << (i - 1))][i - 1]); Max[j][i] = max(Max[j][i - 1], Max[j + (1 << (i - 1))][i - 1]); } } for(int i = 0;i < m; ++ i) { int l,r; cin >> l >> r; int s = Log[r - l + 1]; int ma = max(Max[l][s], Max[r - (1 << s) + 1][s]); int mi = min(Min[l][s], Min[r - (1 << s) + 1][s]); printf("%d\n", ma - mi); } return 0; }
这篇关于算法学习(5):ST表的文章就介绍到这儿,希望我们推荐的文章对大家有所帮助,也希望大家多多支持为之网!
- 2024-07-02springboot项目无法注册到nacos-icode9专业技术文章分享
- 2024-06-26结对编程到底难不难?答案在这里
- 2024-06-19《2023版Java工程师》课程升级公告
- 2024-06-15matplotlib作图不显示3D图,怎么办?
- 2024-06-1503-Loki 日志监控
- 2024-06-1504-让LLM理解知识 -Prompt
- 2024-06-05做软件测试需要懂代码吗?
- 2024-06-0514-ShardingSphere的分布式主键实现
- 2024-06-03为什么以及如何要进行架构设计权衡?
- 2024-05-31全网首发第二弹!软考2024年5月《软件设计师》真题+解析+答案!(11-20题)